Pairing off objects multiple times such that no pairs repeat I am in a class with an even number of students $N$ where we work in pairs for each problem set, and where ideally (in a stylised version of the problem), no student works with someone she has worked with before on a new problem set. I ended up idly wondering the following:


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*How many problems sets can be assigned before at least one student must work with someone he has worked with before? I know how to calculate the total number of possible configurations in the first turn, where no pairings are not permissible.
However, after the first turn, the set of permissible pairings depends on the pairings that were chosen for the first turn in a somewhat trickier way, and I'm also having trouble framing this question properly.

*Suppose we can now work with someone $k \geq 1$ times. Is the maximum number of problem sets now simply $k (\text{number of problem sets when }k=1)$? It seems doubtful to me somehow, but I can't give a satisfactory justification for either result. (Presumably if I had a result for the first question and the underlying intuition, I would have a better idea of things.)
Could someone point me in the right direction for solving this problem?
 A: Case $k=1$:
We're interested in the number of problem sets that can be done with new partners (for every student). Since a student has $N-1$ classmates, we know it's at most $N-1$. 
If we find a way to rotate the students $N-1$ times such that they all have a new partner each time, then we know that $N-1$ is in fact the maximal amount of sheets that can be discussed this way with $N$ students.
Here is such a way: Suppose the students are sitting in a circle and number them $1$ through $N$ in order. For the 1st problem set let them work with their neighbors such that $1$ works with $2$, $3$ with $4$, $\ldots$, and $N-1$ with $N$. For the 2nd sheet let them work with the neighbor on the other side, so now $2$ works with $3$, $4$ with $5$, ..., and $N$ with $1$.
For the 3rd sheet let them each skip one person, so $1$ works with $3$, $2$ with $4$, $5$ with $7$, ... For the 4th sheet, let them again skip one person, but to the other side, so $1$ works with $N-1$, $2$ with $N$, $3$ with $5$, $4$ with $6$ and so on. For problem sheets 5 and 6 each student first works with the person three to the one side (skipping two), then three to the other, and so on. Continue like that, until they work with the person across ($N$ is even, so there is a person directly accross!), so $1$ with $\frac{N}{2}$, $2$ with $\frac{N}{2}+1$, .. Now they have only one possibility, because skipping $\frac{N}{2}-1$ to the left leads to the same partner as skipping $ \frac{N}{2}-1$ to the right.
So when they skip between $0$ and $\frac{N}{2}-2$ classmates, there are two ways to work with new partners (to the left and to the right), and when skipping $\frac{N}{2}-1$ there is one way. Thus with this rotation, we can give out $(\frac{N}{2}-1) \cdot 2 +1 = N-1$ problem sheets, and every student worked with each classmate exactly once.
$N-1$ is our solution!
Case $k>1$: 
Just go through the rotation multiple times..
